Problem: Simplify and expand the following expression: $ \dfrac{4}{r - 2}+ \dfrac{3}{2r - 20}- \dfrac{2}{r^2 - 12r + 20} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the second term: $ \dfrac{3}{2r - 20} = \dfrac{3}{2(r - 10)}$ We can factor the quadratic in the third term: $ \dfrac{2}{r^2 - 12r + 20} = \dfrac{2}{(r - 2)(r - 10)}$ Now we have: $ \dfrac{4}{r - 2}+ \dfrac{3}{2(r - 10)}- \dfrac{2}{(r - 2)(r - 10)} $ The least common multiple of the denominators is: $ (r - 2)(r - 10)$ In order to get the first term over $(r - 2)(r - 10)$ , multiply by $\dfrac{2(r - 10)}{2(r - 10)}$ $ \dfrac{4}{r - 2} \times \dfrac{2(r - 10)}{2(r - 10)} = \dfrac{8(r - 10)}{(r - 2)(r - 10)} $ In order to get the second term over $(r - 2)(r - 10)$ , multiply by $\dfrac{r - 2}{r - 2}$ $ \dfrac{3}{2(r - 10)} \times \dfrac{r - 2}{r - 2} = \dfrac{3(r - 2)}{(r - 2)(r - 10)} $ In order to get the third term over $(r - 2)(r - 10)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{2}{(r - 2)(r - 10)} \times \dfrac{2}{2} = \dfrac{4}{(r - 2)(r - 10)} $ Now we have: $ \dfrac{8(r - 10)}{(r - 2)(r - 10)} + \dfrac{3(r - 2)}{(r - 2)(r - 10)} - \dfrac{4}{(r - 2)(r - 10)} $ $ = \dfrac{ 8(r - 10) + 3(r - 2) - 4} {(r - 2)(r - 10)} $ Expand: $ = \dfrac{8r - 80 + 3r - 6 - 4}{2r^2 - 24r + 40} $ $ = \dfrac{11r - 90}{2r^2 - 24r + 40}$